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I am interested in a method that allows to find global maxima / minima of any elementary function (continuous on their domain) of 2 variables on open sets. If there is no global maxima / minima, then find any of upper or lower bounds of the range of the function. If they don't exist also, prove that.
I am not interested in specific solutions that can be conveniently applicable to a specific example below. I'm interested in formally described method (if the method does exist), applicable to all of them. I'll be glad for any links to scientific literature. Will be also glad if the method can be generalized to functions of more that 2 variables.

Examples just for better understanding. Find global maxima and minima, or any of lower or upper bounds of

  1. f(x;y) = xy on R²,
  2. f(x;y) = 1/(xy) on the domain,
  3. f(x;y) = x² + 2x + 3y² where x² + y² < 1
  4. f(x;y) = log₂(x) + logₓ(y) + log_y(8) where x,y > 1 (global minima here is ∛81)

    or prove that they don't exist.
Noname Nonameovich
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  • Such a method exists for functions of 1 variable. After finding local extremas, you'll need to finding limits to endpoints of open intervals or limits to ±∞. https://math.stackexchange.com/questions/1598198/steps-of-finding-an-absolute-extremum-on-an-open-interval – Noname Nonameovich Nov 30 '22 at 22:33
  • For any continuous function on an open set, if the second partial derivative test finds no critical points, or that every critical point is a saddle, then there is no global minimum or maximum. But it seems maybe you actually want to find the infimum and supremum, instead of the global minimum and maximum? – aschepler Nov 30 '22 at 23:49
  • Oh, I didn't think about it. Indeed, a global minimum is a local minimum, and if there is no local minimum then there is no global one also. :facepalm: Then yes, I would like to find infimum and supremum, and if they don't exist, then any lower or upper bound, and if they don't exists, prove that. – Noname Nonameovich Nov 30 '22 at 23:58
  • I'll edit the question. Thank you. – Noname Nonameovich Nov 30 '22 at 23:58
  • Though wait. How can I formally prove that for a function of 2 variables on an open set, a global maximum is always a critical point? It may be not true for a closed set. For example f(x) = x² on [-1;1] has the only critical point x=0, but also has a global matima at x=1, which is not a critical point. – Noname Nonameovich Dec 01 '22 at 00:16
  • That comes from the definition of a partial derivative, since an open neighborhood of the point will contain points in both directions along that line. – aschepler Dec 01 '22 at 00:20
  • For now even if the the second derivative test gives us local minima and maxima points, we don't know any way to prove or disprove that they are global maxima and minima points. – Noname Nonameovich Dec 01 '22 at 01:32

1 Answers1

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The steps to investigating global extrema of smooth functions on open sets are:

  • Find all the local extrema.
  • Examine the behavior of the function outside compact sets - can you remove a compact set such that the function is bounded (above or below)? What is the infimum/supremum? If the function is unbounded in the corresponding direction, the infimum is $-\infty$, the supremum is $+\infty$.
  • If the function is unbounded in a direction, then it has no global extremum in that direction. Otherwise,
  • If any of the local minimums is less than the infimum, then the lowest is the global minimum. Otherwise the infimum is global (but not a minimum). If any of the local maximums is greater than the supremum, then the highest is the global maximum, otherwise the supremum is global.

You can interpret the "behavior outside compact sets" to mean the limits along curves leaving the domain at some point, or going off to infinity in some fashion.

In your examples,

  1. Clearly, $\lim_{x \to \pm \infty} f(x, 1) = \pm\infty$, so this function is unbounded in both directions.
  2. Equally clearly, $\lim_{x\to 0\pm} f(x,1) = \pm\infty$, also unbounded.
  3. This function extends continuously to the closure of the given domain. The behavior of $f$ outside compact sets is exactly the behavior on that boundary. So you handle this by solving the problem on the closed disk. The only difference is if the extremum occurs only on the boundary it is an infimum/supremum, not a maximum, minimum.
  4. Each term is bounded below by $0$, it is not unbounded in that direction. In the other direction, if $x \to \infty$, so does the first term, and as they are bounded below, the other two terms cannot compensate, so the entire function goes to $\infty$ regardless of what $y$ does. If $x$ does not go to $\infty$, then the second term goes to $\infty$ with $y$, and if $y \to 1$, the third term goes to $\infty$ regardless of what $x$ does. Finally if $y$ does not go to $1$, the second term goes to $\infty$ as $x \to 1$. So on all the "boundaries", the function goes to $\infty$. The lowest local minimum will be the global minimum.
Paul Sinclair
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  • Thanks for the ideas. Your method is pretty intuitive, though it will be tough to base is formally. – Noname Nonameovich Dec 02 '22 at 01:19
  • Let's define f(x;y)=x²+y²+xy+x on R². lim f(x;y) = +∞ to all boundaries (x->±∞ and y->±∞). The only local extrema is local minimum point (-2/3;1/3), f(-2/3;1/3) = -1/3. If you see the graph of the function https://www.geogebra.org/3d/jpevf9cp it becomes obvious that (-2/3;1/3) is also a global minima. The problem is that I don't know any ways (except looking at the graph an intuitively realize it) to disprove if someone states that there exists a point (xₒ;yₒ) such that for example f(xₒ;yₒ) = -10 (or just < -1/3 in general). – Noname Nonameovich Dec 02 '22 at 01:19
  • It is not that tough to base formally. The only part I fudged was simply because I didn't want to bog down the discussion with limits of nets of sets, a concept which has long been fully defined. As for proving that such an $(x_0, y_0)$ is impossible, let $\gamma(t)$ be the line connecting $\gamma(0) = (-2/3, 1/3)$ to $\gamma(1) = (x_0, y_0)$. The only critical point of $f\circ \gamma$ is at $0$, and $f\circ \gamma$ increases in either direction as you move away from $0$. – Paul Sinclair Dec 02 '22 at 13:03